1) Propositional functions are propositions that contain variables and have no truth value until the variables are assigned values or quantified. 2) Quantifiers like "for all" (universal quantification) and "there exists" (existential quantification) are used to bind variables and give propositional functions truth values. 3) Quantification can be thought of as nested loops over variables, with universal quantification checking for truth at each value and existential checking for at least one true value.

1. Rosen 1.3

2. Propositional Functions
• Propositional functions (or predicates) are
propositions that contain variables.
• Ex: Let P(x) denote x > 3
• P(x) has no truth value until the variable x is
bound by either
– assigning it a value or by
– quantifying it.

3. Assignment of values
Let Q(x,y) denote “x + y = 7”.
Each of the following can be determined as T or F.
Q(4,3)
Q(3,2)
Q(4,3)  Q(3,2)
[Q(4,3)  Q(3,2)]

4. Quantifiers
Universe of Discourse, U: The domain of a variable in
a propositional function.
Universal Quantification of P(x) is the
proposition:“P(x) is true for all values of x in U.”
Existential Quantification of P(x) is the proposition:
“There exists an element, x, in U such that P(x) is
true.”

5. Universal Quantification of P(x)
xP(x)
“for all x P(x)”
“for every x P(x)”
Defined as:
P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U
Example:
Let P(x) denote x2  x
If U is x such that 0 < x < 1 then xP(x) is false.
If U is x such that 1 < x then xP(x) is true.

6. Existential Quantification of P(x)
xP(x)
“there is an x such that P(x)”
“there is at least one x such that P(x)”
“there exists at least one x such that P(x)”
Defined as:
P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U
Example:
Let P(x) denote x2  x
If U is x such that 0 < x  1 then xP(x) is true.
If U is x such that x < 1 then xP(x) is true.

7. Quantifiers
xP(x)
•True when P(x) is true for every x.
•False if there is an x for which P(x) is false.
xP(x)
•True if there exists an x for which P(x) is true.
•False if P(x) is false for every x.

8. Negation (it is not the case)
xP(x) equivalent to xP(x)
•True when P(x) is false for every x
•False if there is an x for which P(x) is true.
 xP(x) is equivalent to xP(x)
•True if there exists an x for which P(x) is false.
•False if P(x) is true for every x.

9. Examples 2a
Let T(a,b) denote the propositional function “a
trusts b.” Let U be the set of all people in the
world.
Everybody trusts Bob.
xT(x,Bob)
Could also say: xU T(x,Bob)
 denotes membership
Bob trusts somebody.
xT(Bob,x)

10. Examples 2b
Alice trusts herself.
T(Alice, Alice)
Alice trusts nobody.
x T(Alice,x)
Carol trusts everyone trusted by David.
x(T(David,x)  T(Carol,x))
Everyone trusts somebody.
x y T(x,y)

11. Examples 2c
x y T(x,y)
Someone trusts everybody.
y x T(x,y)
Somebody is trusted by everybody.
Bob trusts only Alice.
x (x=Alice  T(Bob,x))

12. Bob trusts only Alice.
x (x=Alice  T(Bob,x))
Let p be “x=Alice”
q be “Bob trusts x”
p q p  q
T T T
T F F
F T F
F F T
True only when
Bob trusts Alice
or Bob does not
trust someone
who is not Alice

13. Quantification of Two Variables
(read left to right)
xyP(x,y) or yxP(x,y)
•True when P(x,y) is true for every pair x,y.
•False if there is a pair x,y for which P(x,y) is false.
xyP(x,y) or yxP(x,y)
True if there is a pair x,y for which P(x,y) is true.
False if P(x,y) is false for every pair x,y.

14. Quantification of Two Variables
xyP(x,y)
•True when for every x there is a y for which P(x,y) is true.
(in this case y can depend on x)
•False if there is an x such that P(x,y) is false for every y.
yxP(x,y)
•True if there is a y for which P(x,y) is true for every x.
(i.e., true for a particular y regardless (or independent) of x)
•False if for every y there is an x for which P(x,y) is false.
Note that order matters here
In particular, if yxP(x,y) is true, then xyP(x,y) is true.
However, if xyP(x,y) is true, it is not necessary that yxP(x,y)
is true.

15. Examples 3a
Let L(x,y) be the statement “x loves y” where U for both
x and y is the set of all people in the world.
Everybody loves Jerry.
xL(x,Jerry)
Everybody loves somebody.
x yL(x,y)
There is somebody whom everybody loves.
yxL(x,y)

16. Examples 3b1
There is somebody whom Lydia does not love.
xL(Lydia,x)
Nobody loves everybody. (For each person there is at
least one person they do not love.)
xyL(x,y)
There is somebody (one or more) whom nobody loves
y x L(x,y)

17. Examples 3b2
There is exactly one person whom everybody loves.
xyL(y,x)?
No. There could be more than one person everybody
loves
x{yL(y,x)  w[(yL(y,w))  w=x]}
If there are, say, two values x1 and x2 (or more) for
which L(y,x) is true, the proposition is false.
x{yL(y,x)  w[(yL(y,w))  w=x]}?
xw[(y L(y,w))  w=x]?

18. Examples 3c
There are exactly two people whom Lynn loves.
x y{xy  L(Lynn,x)  L(Lynn,y)}?
No.
x y{xy  L(Lynn,x)  L(Lynn,y)  z[L(Lynn,z)
(z=x  z=y)]}
Everyone loves himself or herself.
xL(x,x)
There is someone who loves no one besides himself or
herself.
xy(L(x,y)  x=y)

19. Thinking of Quantification as
Loops
Quantifications of more than one variable
can be thought of as nested loops.
•For example, xyP(x,y) can be thought
of as a loop over x, inside of which we
loop over y (i.e., for each value of x).
• Likewise, xyP(x,y) can be thought of
as a loop over x with a loop over y nested
inside. This can be extended to any
number of variables.

20. Quantification as Loops
Using this procedure
•xyP(x,y) is true if P(x,y) is true for all values of x,y
as we loop through y for each value of x.
•xyP(x,y) is true if P(x,y) is true for at least one set of
values x,y as we loop through y for each value of x.
…And so on….

21. Quantification of 3 Variables
Let Q(x,y,z) be the statement “x + y = z”, where x,y,z
are real numbers.
What is the truth values of
•xyzQ(x,y,z)?
•zxyQ(x,y,z)?

22. Quantification of 3 Variables
Let Q(x,y,z) be the statement “x + y = z”, where x,y,z
are real numbers.
•xyzQ(x,y,z)
is the statement, “For all real numbers x and for all
real numbers y, there is a real number z such that
x + y = z.”
True

23. Quantification of 3 Variables
Let Q(x,y,z) be the statement “x + y = z”, where x,y,z
are real numbers.
zxyQ(x,y,z)
is the statement, “There is a real number z such that
for all real numbers x and for all real numbers y,
x + y = z.”
False

24. Examples 4a
Let
P(x) be the statement: “x is a Georgia Tech student”
Q(x) be the statement: “ x is ignorant”
R(x) be the statement: “x wears red”
and U is the set of all people.
No Georgia Tech students are ignorant.
x(P(x) Q(x))
x(P(x) Q(x))
OK by Implication equivalence.
x(P(x)  Q(x))
Does not work. Why?

25. Examples 4a
x(P(x)  Q(x))
x (P(x)  Q(x)) Negation equivalence
x ( P(x)  Q(x)) Implication equivalence
x (  P(x)   Q(x)) DeMorgans
x ( P(x)   Q(x)) Double negation
Only true if everyone is a GT student and is not ignorant.
No Georgia Tech students are ignorant.
x(P(x) Q(x))

26. Examples 4a
P(x) be the statement: “x is a Georgia Tech student”
Q(x) be the statement: “ x is ignorant”
R(x) be the statement: “x wears red”
and U is the set of all people.
No Georgia Tech students are ignorant.
x(P(x)  Q(x))
Also works. Why?

27. Examples 4a
x(P(x)  Q(x))
 x (P(x)  Q(x)) Negation equivalence
x (P(x)  Q(x)) DeMorgan
x (P(x) Q(x)) Implication equivalence
No Georgia Tech students are ignorant.
x(P(x) Q(x))

28. Examples 4b
Let
P(x) be the statement: “x is a Georgia Tech student”
Q(x) be the statement: “ x is ignorant”
R(x) be the statement: “x wears red”
and U is the set of all people.
All ignorant people wear red.
x(Q(x) R(x))

29. Examples 4c
Let
P(x) be the statement: “x is a Georgia Tech student”
Q(x) be the statement: “ x is ignorant”
R(x) be the statement: “x wears red”
and U is the set of all people.
No Georgia Tech student wears red.
x(P(x) R(x))
What about this?
x(R(x)  P(x))

30. Examples 4d
If “no Georgia Tech students are ignorant” and “all
ignorant people wear red”, does it follow that “no
Georgia Tech student wears red?”
x((P(x) Q(x))  (Q(x) R(x)))
NO
Some misguided GT student might wear red!!
This can be shown with a truth table or Wenn diagrams